1. Find the magnitude of the resultant of the two concurrent forces F1 and F2 in Figure 1. 2. In Question 1 above, determine the angle that the resultant makes with the force F1. 3... 1. Find the magnitude of the resultant of the two concurrent forces F1 and F2 in Figure 1. 2. In Question 1 above, determine the angle that the resultant makes with the force F1. 3. Determine the angle \(\Theta\) if the resultant of the two concurrent forces F1 and F2 in Figure 2 is 200 N. Read more

Steps to Solve

1. Find the x and y components of forces F1 and F2

F1 has a magnitude of 110 N and is directed along the negative x-axis, so its x-component is -110 N and its y-component is 0 N. F2 has a magnitude of 55 N and acts at an angle of 80 degrees with respect to the negative x-axis. $F_{2x} = 55 \cdot \cos(80^\circ)$ $F_{2x} = -9.55N$ $F_{2y} = 55 \cdot \sin(80^\circ)$ $F_{2y} = 54.16N$

2. Calculate the x and y components of the resultant force

Add the x-components of F1 and F2 to get the x-component of the resultant force $F_x$. $F_x = F_{1x} + F_{2x} = -110 + (-9.55) = -119.55 N$

Add the y-components of F1 and F2 to get the y-component of the resultant force $F_y$. $F_y = F_{1y} + F_{2y} = 0 + 54.16 = 54.16 N$

3. Calculate the magnitude of the resultant force

Use the Pythagorean theorem to find the magnitude of the resultant force $F$. $F = \sqrt{F_x^2 + F_y^2}$ $F = \sqrt{(-119.55)^2 + (54.16)^2}$ $F = \sqrt{14292.2025 + 2933.3056}$ $F = \sqrt{17225.5081}$ $F = 131.25 N$

4. Determine the angle that the resultant makes with the force $F_1$

Calculate the angle $\theta$ that the resultant force makes with the negative x-axis: $\theta = \arctan(\frac{F_y}{F_x})$ $\theta = \arctan(\frac{54.16}{-119.55})$ $\theta = -24.35^\circ$

Since $F_1$ is along the negative x-axis, the angle that the resultant makes with $F_1$ is the absolute value of $\theta$: $|\theta| = |-24.35^\circ| = 24.35^\circ \approx 24.4^\circ$

5. Determine the angle Θ if the resultant of the two concurrent forces $F_1$ and $F_2$ is 200 N

Let the resultant force be $R = 200$ N. We are given $F_1 = 100$ N and $F_2 = 80$ N. Also, let the angle between $F_1$ and $F_2$ be $\Theta$. We can use the law of cosines to find the angle $\Theta$: $R^2 = F_1^2 + F_2^2 + 2 \cdot F_1 \cdot F_2 \cdot \cos(\Theta)$ In this case, we should use the following equation: $R^2 = F_1^2 + F_2^2 - 2 \cdot F_1 \cdot F_2 \cdot \cos(\Theta)$ $200^2 = 100^2 + 80^2 - 2 \cdot 100 \cdot 80 \cdot \cos(\Theta)$ $40000 = 10000 + 6400 - 16000 \cdot \cos(\Theta)$ $40000 = 16400 - 16000 \cdot \cos(\Theta)$ $23600 = -16000 \cdot \cos(\Theta)$ $\cos(\Theta) = \frac{23600}{-16000} = -1.475$ This equation is not possible since the $\cos(\Theta)$ must be between -1 and 1. There might be a typo, and $R$ could be the angle between $F_1$ and the resultant force. Let us consider that interpretation.

The law of cosines to find the angle $\alpha$ between $F_1$ and the resultant force $R$: $F_2^2 = F_1^2 + R^2 - 2 \cdot F_1 \cdot R \cdot \cos(\alpha)$ $80^2 = 100^2 + 200^2 - 2 \cdot 100 \cdot 200 \cdot \cos(\alpha)$ $6400 = 10000 + 40000 - 40000 \cdot \cos(\alpha)$ $6400 = 50000 - 40000 \cdot \cos(\alpha)$ $-43600 = -40000 \cdot \cos(\alpha)$ $\cos(\alpha) = \frac{43600}{40000} = 1.09$ Unfortunately, we also run into issues with this situation as the value of $\cos{\alpha}$ should be between $-1$ and $1$. Therefore, the question is not solvable based on the assumption that the angle, $\Theta$, corresponds to the angle between $F_1$ and $F_2$.